Problem: You have found the following ages (in years) of all 6 turtles at your local zoo: $ 4,\enspace 50,\enspace 63,\enspace 2,\enspace 82,\enspace 99$ What is the average age of the turtles at your zoo? What is the variance? You may round your answers to the nearest tenth.
Explanation: Because we have data for all 6 turtles at the zoo, we are able to calculate the population mean $({\mu})$ and population variance $({\sigma^2})$ To find the population mean , add up the values of all $6$ ages and divide by $6$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{6}} x_i}{{6}} $ $ {\mu} = \dfrac{4 + 50 + 63 + 2 + 82 + 99}{{6}} = {50\text{ years old}} $ Find the squared deviations from the mean for each turtle. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $4$ years $-46$ years $2116$ years $^2$ $50$ years $0$ years $0$ years $^2$ $63$ years $13$ years $169$ years $^2$ $2$ years $-48$ years $2304$ years $^2$ $82$ years $32$ years $1024$ years $^2$ $99$ years $49$ years $2401$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{2116} + {0} + {169} + {2304} + {1024} + {2401}} {{6}} $ $ {\sigma^2} = \dfrac{{8014}}{{6}} = {1335.67\text{ years}^2} $ The average turtle at the zoo is 50 years old. The population variance is 1335.67 years $^2$.